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Say the vector space $V$ has two bases, $B$ and $B'$. There exists a matrix $P$ such that $BP=B'$.

$B$ is of the form $\begin{pmatrix} v_1&v_2&\dots&v_n\\ \vdots&\vdots&\vdots&\vdots\end{pmatrix}$, and $B'$ is of the form $\begin{pmatrix} v'_1&v'_2&\dots&v'_n\\ \vdots&\vdots&\vdots&\vdots\end{pmatrix}$.

Let $P$ be $\begin{pmatrix} p_{11}&p_{12}&\dots&p_{1n}\\ \vdots&\vdots&\vdots&\vdots\end{pmatrix}$. This means that $p_{11}v_1+p_{21}v_2+\dots p_{n1}v_n=v'_1$.

I have read in some places that $T(c_1v_1+c_2v_2+\dots c_nv_n)=\begin{pmatrix} p_{11}&p_{12}&\dots&p_{1n}\\ \vdots&\vdots&\vdots&\vdots\end{pmatrix}\begin{pmatrix}c_1\\\vdots\\c_n\end{pmatrix}$.

But isn't that just $c_1v'_1+c_2v'_2+\dots+c_nv'_n$ expressed in terms of $v_1,v_2,\dots,v_n$?

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  • $\begingroup$ Presumably $B,B'$ have the form $v_{1k}, v_{1k}'$? What is $T$? $\endgroup$ – copper.hat Aug 16 '14 at 18:05
  • $\begingroup$ Yeah, dude, what is T? $\endgroup$ – shooting-squirrel Aug 16 '14 at 18:07
  • $\begingroup$ @copper.hat- Isn't $T$ equal to $P$? The transformation in general? $\endgroup$ – fierydemon Aug 16 '14 at 18:11
  • $\begingroup$ I don't know, you haven't said what $T$ is... $\endgroup$ – copper.hat Aug 16 '14 at 18:21

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